iGCSE Physics (0625) 1.5.2 Turning effect Paper 4 -Exam Style Questions- New Syllabus

(a)
For the correct answer:
The product of the force and the perpendicular distance from the pivot.

In physics, the moment of a force is a measure of its turning effect. It is quantitatively defined as the force multiplied by the perpendicular distance between the line of action of the force and the point of rotation (the pivot).

(b)
For the correct answer:
Mark “X” at the $50\text{ cm}$ mark on the ruler.

Since the ruler is “uniform,” its centre of gravity is at its geometric centre. For a $100\text{ cm}$ ruler, the midpoint is at $50\text{ cm}$. This is to the right of the pivot ($42\text{ cm}$) and to the left of the $0.12\text{ N}$ weight ($80\text{ cm}$).

(c)(i)
For the correct answer:
$m = 0.081\text{ kg}$ (shown via moments)

Using Principle of Moments: $\text{Clockwise} = \text{Anticlockwise}$.
Distances from pivot: $0.34\text{ N}$ is $32\text{ cm}$ ($42-10$), $0.12\text{ N}$ is $38\text{ cm}$ ($80-42$), and ruler weight $W$ is $8\text{ cm}$ ($50-42$).
$(W \times 8) + (0.12 \times 38) = (0.34 \times 32) \Rightarrow 8W + 4.56 = 10.88$.
$W = 0.79\text{ N} \Rightarrow m = \frac{0.79}{9.8} \approx 0.081\text{ kg}$.

(c)(ii)
For the correct answer:
$520\text{ kg/m}^3$

Volume $V = 1.0 \times (2.6 \times 10^{-2}) \times (6.0 \times 10^{-3}) = 1.56 \times 10^{-4}\text{ m}^3$.
Density $\rho = \frac{m}{V} = \frac{0.081}{1.56 \times 10^{-4}} \approx 519.23\text{ kg/m}^3$. Rounded to two significant figures: $520\text{ kg/m}^3$.